HW11 - Problem 9.14 quuation 9.73 is replaced by Ems +...

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Unformatted text preview: Problem 9.14 quuation 9.73 is replaced by Ems + Eons}; = EDTfiTLand Eq. 9.39 becomes Emir h Egan X en) = fifiihfi x fir). The y component of the first equation is EUR sin 93 = EDT sin ET; the a: component of the second is Eon sin 9;; = —,6EDT sitter. Comparing these two. we conclude that sin 93 = sin 97- : 0, and hence 83 = 9-1- = O. qed Problem 9.15 146‘" + Be‘“ 2 Ce'CI for all 1:, so (using 5 = O). A + B 2 C. Differentiate: iaAe'" + ‘ilJBe'rm = icCe'”, so (using a: = 0), ml + bB = cC. Difl'erentiate again: —a:2Ae'” — bEBe'b” = —c2Ce"", so (using 1' = 0). G214 + sz = c202 02A + (:33 = C(cC) = C(aA + 53); (A + B)(a"‘A + 523) = (A + B)c(a.A + 53) = cC(aA + bB); aw + I;er + (FAB + 1332 = (aA + as)2 = (EA? + 2abAB + b232, or (a2 + b? — 2ab)AB = u. or {a — b}2AB = 0. But. A and B are nonzero. so a = 5. Therefore (A + B)e‘” : Ce‘”. a(A + B) = (:0, or :10 = c0, so (since 0 9t 0) a = c. Gentleman.- a. = b = c. qed Problem 9.16 u {m = new-we. } B; = $Eore‘lk""“")(—co591 it + sing] fi); 1 ER E1 R3109; r—wtly, fin v—Eone'lkR‘PWHCOS 91 i + sin 91 i); 1 { ET EDT esfltg- rut-It)?" } 1 - v—Eure‘(k"'"_”" (— (:06 92 i + sin 01 i); 2 DJ .3 || (i) ElElJ‘ = EzEé', (iii) E',‘ = Ell, Boundary conditions: (11) 31* = 52*, (iv) filial} = “Agni. , sin 02 112 La .' = — w of refraction Sin 91 vi exponential factors in applying the boundary conditions] Boundary condition (i): D = 0 (trivial). Boundary condition (iii): E"), + Eon = 30?. 1 - - 1 - - - Boundary condition (ii): v—Eol sin 91 + ULEDR sinal = 11—13% sin 92 :> En, 1» EUR = ( 1 1 2 . [N0te: k;-r—wt=kR-r—wt=kT-r—wt.atz=0,sowecandropall 'U; sin 92 - '02 sin 91 ) EDT- But the term in parentheses is 1, by the law of refraction, so this is the same as (ii). 1 Boundary condition (iv): — [ifihfi— cos 9;) + :Jl—Enn cos 91] = if)“. (7 cos 9;) => 1 1-51 ' ” “WW2” -°°“” -””” E a : ~———fi .L 2 - z . Th - : . 2 Solving for Eon and EDT: 2E1), = (1 + (1,8)17/‘0, => Eur = (1 +Qfi) Em; - — - 2 1 + (13 ~ ~ 1 — a6 - = — = — E = - E0“ E0? E0‘ (1+afi 1+afi) “.4130” (1+a5) Eo’ Since a and ,6 are positive, it follows that 2/ (1 + 0:3) is positive, and hence the transmitted Wave is in phase ‘2 . 1 + 043) Em. The reflected wave IS with the incident wave. and the (real) amplitudes are related by E9, = ( 1/1~sin2B/fi2 _ 5275i“ 9 , where 9 is the angle ofincidenoe, cos 6 cos 6' To construct the graphs, note that 9.6 = fl V2.25 — sin) 6 sn.f0rt3:15.afi= cosB r ['31 D' 10 20 .50 4” 50 60 7U S0 90 BI Is there a Brewster's angle? Well, Eon = 0 would mean that (1,6 = 1, and hence that. ‘/1— v2 2; 2sin29 2 2 ( f1) gzfl, 011- sin29= 00329. 30 a = — = cost? 1); #101 2 1 = [sin2 0 + (tag/n1)2 0052 9]. Since in a: #2, this means 1 z (cg/‘01P. which is only true for optically 1 indistinguishable media, in which case there is of course no reflection—but that would be true at any angle, not just at a special “Brewster’s angle”. [if p; were substantially different from in, and the relative velocities were just right, it would be possible to get a Brewster's angle for this case, at 2 2 '01 2 #2 2 2 (‘vll'vzl2 -1 (me/#16:) " 1 82/61) - (fir/11c) — = 1 — 9 — H = — = —-—-—-—-.- = —— l l “’5 + l ) °°S =” “’5 9 Cruz/#02 — 1 (rm/m)” - 1 (Its/.01) — (it/M But the media would be very peculiar.] By the same token, 63 is either always 0, or always 1r, for a given interface—it does not switch over as you change 9, the way it does for polarization in the plane of incidence. In particular, if ,8 = 3/2, then at} > 1, for «2.25 - sin2 9 (2,8: €066 >1if2.25—sin99>cosefl, or2.25>sin29+cos"9=1..f In general, for 6 >1, 0:6 > 1, and hence (in = 1r. For [3 < 1. :13 <1. and «in = D. At normal :nczdcnce, or = 1, so Fresnel’s equations reduce to E0, = Em; Eon = %|Eon consistent with Eq. 9.82. 2 2 Reflection and Transmission coefficients: R = (Eon) = (I ~ ale) . Referring to Eq. 9.116, En, 1+afi 2 2 ca Ear _ 2 T7€1fl1a(Eol) 7 flat-Fang) - (1 —a,6)2 +4as _ 1 —2c:,6+c:2,62 +4043 7 (1+afi)2 i (1 + and)2 _ (1 +a,(3)2 _ (1 +045?)2 _ R+T= 1.! Problem 9.30 Following Sect. 9.5.2, the problem is to solve Eq. 9.181 with E, 95 0,8, = 0, subject to the boundary conditions 9.175. Let Eating) = X(:r)l’(y); as before. we obtain X(:c) = Asin(k;:r) + Bcos[kz:r). But the boundary condition requires E, = 0 (and hence X = 0) when a: = 0 and a: = a. so B = 0 and k, = mrr/a. But this time 111 = 1,2, 3, . .. , but not zero. since m = 0 would kill X entirely. The same goes for Y(y). Thus E, = Ensin sin (“7:”) with n..m= 1.2,3..... The rest is the same as for TB waves: w...“ = elm/(rat/a.)2 + (rt/(7)2 is the cutoff frequency, the wave velocity is 'u = ch/l -— (Lawn/(JP, and the group velocity is v9 = cv’l — lwmnlw)? The lowest TM mode is 11, with cutoff frequency um = can/(l/a)2 + (1/1))”. So the ratio of the lowat TM frequency to the lowest TE frequency is M— W = x/l + (ca/b 2. (car/a) Problem 9.31 6E 8E k . (al vs: gages.) =0 (zv-B = 5233.5) 2 0/;VxE: a; L: a; 2 = —wr$; 3%? : —%—Sm(k:_ wt) 43 if (since k = w/c);V x B = "6652" 51+ %§;(ng)2 = %s_mlk:_ wt) § ; i8_E _ E sinUcz —wt) c2 at — c2 s (b) To determine A, use Gauss's law for a cylinder of radius a and length dz: fE - do = Qwenfldz = 61099“ = Elhdz => A = 21rcgEu cosUtz — wt). 0 0 To determine I, use Ampére’s law for a circle of radius 3 (note that the displacement current through this loop is zero. since Eis in the s direction): fB-dl : %w(2ns) = null.“ => I = 2:05;" cosUcz — wt). The charge and current on the outer conductor are precisely the of these, since E = B = 0 inszde the metal. and hence the total enclosed charge and current must be zero. § J. Boundary conditions: E“ = E, = 0 (:BJ- = B, = 0 ,/. Problem 10.3 3A 1 :1. 7_VV_——._ —— . B—VXA— E at 4602" 1 This is a funny set of potentials for a stationary point charge (,1 at the origin. [V = mg, A z 0 would. of course, be the customary choice.) Evidently p = (163(1); J = 0. Problem 10.10 he Renew/(telekah $1 a. 4x a. Mr 4 41r IL c bda: :‘B Butforthe complete loop,fdl=fl,soA=M{lfdl+lfdl+2if 411' a, 1 b 3 a }. Here [1 cl] = 2a.)? (inner circle), [2:11 = —2bi (outer circle), so duo—"’3 l l_ . _go£§ . A— 4” [0(2a)+b( 2b)+21n[b/a]]x=> A— 2" ln(b/a)x, E The changing magnetic field induces the electric field. Since we only know A at one paint (the center), we can't compute V x A to get B. 6A_ A: - —E — 2x x. ...
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This note was uploaded on 12/29/2011 for the course PHYSICS 411 taught by Professor Agshe during the Summer '11 term at Maryland.

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HW11 - Problem 9.14 quuation 9.73 is replaced by Ems +...

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